How do I solve this polynomial equation in $x$ and $y$?

Question

$$x^2 + y^2 - 2xy - 6x - 6y + 9 = 0$$

How can I solve this equation to obtain values of $x$ and $y$ that are solution to this polynomial?

Answer 1

Choose a value for x, pretty much arbitrarily. The result will be a quadratic equation for y. You can use the quadratic formula to determine if there is such a y and, if so, find it. For example, taking x= 0, we have $y^2- 6y+ 9= (y- 3)^2= 0$ so (0, 3) is an (x, y) pair. Taking x= 1, we have $1+ y^2- 2y- 6- 6y+ 9= y^2- 8y+ 3= 0$. By the quadratic formula, $y= \frac{8\pm\sqrt{64- 12}}{2}= \frac{8\pm\sqrt{52}}{2}= 4\pm\sqrt{13}$ so two more pairs are $(1, 4+ \sqrt{13})$ and $(1, 4- \sqrt{13})$.

Answer 2

The polynomial turns out to be

$$(x-y)^2-6(x+y)+9$$ which can be written

$$u^2-6v+9=0.$$

Choosing a value for $u$, you get $$v=\frac{u^2+9}6.$$

Then solve

$$\begin {cases}x-y=u,\\x+y=v.\end{cases}$$

---

In other terms, the curve has the parametric equations

$$\begin {cases}x=\dfrac{u^2+6u+9}{12},\\y=\dfrac{u^2-6u+9}{12}.\end{cases}$$

From the $(u,v)$ representation, which differs from the $(x,y)$ representation by a similarity transform, we know that this is a parabola.